$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Inn{Inn}$Let $S_\infty$ be the group of all permutations of a countable infinite set (enriched with the product Polish topology).
Does there exist a (continuous) group embedding $\phi: S_\infty\to S_\infty$ such that $\phi(G)$ is conjugate to $\phi(H)$ for every pair of (closed) isomorphic groups $G, H\leq S_\infty$?
Note that two subgroups $A, B$ of a group $G$ are conjugate in $G$ if there is $g\in G$ for which $A=gBg^{-1}$.
Additional note: It would be stil of interest for me, if there is an embedding $\phi: S_\infty \to T$ for some Polish group $T$, such that $\phi(G)$ is conjugate to $\phi(H)$ in $T$ for every pair of (closed) isomorphic groups $G, H\leq S_\infty$?
My first guess was to consider $T$ to be the semidirect product of $S_\infty$ with $\Aut(S_\infty)=\Inn(S_\infty)=S_\infty$.
Best Answer
No, there's no such endomorphism. More generally:
Note: every group endomorphism of $S_\omega$ is continuous (automatic continuity, here essentially due to Dixon-Neumann-Thomas: see Prop 4.3 and Thm 4.4 in Rosendal's survey). [The notation $S_\infty$, which implicitly assumes that every infinite set is countable (Cantor disproved this) is inconvenient.]
Let $c$ be any nontrivial permutation of $X$ consisting only of infinite cycles (at least one) and fixed points. Let $c'$ be any permutation of $X$ with both at least one infinite cycle and finite cycles of all sizes.
Since both $c$ and $c'$ have an infinite cycle, the cyclic subgroups $\langle c\rangle$ and $\langle c'\rangle$ are infinite and discrete (hence closed).
(This immediately implies the first proposition, in a strong sense since the embedding property fails for a given pair of isomorphic closed subgroups.)
Lemma Let $H$ be an open subgroup of $S_X$. Then there exists a finite subset $F$ of $X$ such that $H$ is trapped between the pointwise and global stabilizer of $F$. $\Box$
[This is classical. The main case is when $H$ is transitive, in which case the conclusion is $H=S_X$. The general case easily follows.]
Through $f$, we can view $Y$ as an $S_X$-set. For both statements it enough to assume that $Y$ is a transitive $S_X$-set (through $f$). Hence $Y=S_X/H$ for some open subgroup $H$.
For the statement about $c$ (no nonsingleton finite cycle), we can use the lemma and find an open finite index subgroup $H'$ of $H$, such that $H'$ is the pointwise stabilizer of a finite subset. We can identify $S_X/H'$ with the set of pairwise distinct $n$-tuples in $X^n$ for the product action, for some $n\ge 0$. Indeed we see that $c$ has no nonsingleton finite cycle on $\omega^n$. Since the equivariant surjection $S_X/H'\to S_X/H$ is finite-to-one, we deduce that $c$ has also no nonsingleton finite cycle on $S_X/H$.
For the statement about $c'$ (existence of unbounded finite cycles), we can use the lemma again, now defining $H'$ as finite index overgroup of $H$, such that $H'$ is the global stabilizer of a finite subset $F$. If $F$ is empty, we have $H=H'=S_X$ and the statement is void. Assume now $F$ nonempty. We can identify $S_X/H'$ with the set of $n$-elements subsets of $X$ for some $n\ge 1$. Since $c'$ admits a $kn$-cycle for every $k\ge 2$, we see that $c'$ has a $k$-cycle for its action on $S_X/H'$. Hence $c'$ has a $kk'$-cycle for some $k'\ge 1$ on $S_X/H$.
Using automatic continuity, let me mention, out of curiosity:
(Indeed, finding groups of at most continuum cardinal with no faithful action on any countable set has been an open problem long ago, and while various examples are now known, such a construction seems to be a new one. Note that no group topology is involved in the above statement.)
This argument doesn't address general Polish groups as targets. However it suggests that it already makes sense to attempt to make discrete subgroups conjugate, even infinite cyclic ones.